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Chapter 1: Q. 39 (page 135)

Calculate each of the limits in Exercises $29 - 70$ .
$\lim_{x 鈫� \frac{1}{2}} \frac{4 x - 2}{6 x^{2} + x - 2}$ .

Short Answer

Expert verified

The value of $\lim_{x 鈫� \frac{1}{2}} \frac{4 x - 2}{6 x^{2} + x - 2}$ is, $\frac{4}{7}$ .

Step by step solution

01

Step 1. Given information

$\lim_{x 鈫� \frac{1}{2}} \frac{4 x - 2}{6 x^{2} + x - 2}$ .

02

Step 2. The limit is calculated as below:

$\lim_{x 鈫� \frac{1}{2}} \frac{4 x - 2}{6 x^{2} + x - 2} = \lim_{x 鈫� 1 / 2} \frac{4 x - 2}{6 x^{2} + 4 x - 3 x - 2} = \lim_{x 鈫� 1 / 2} \frac{2 (2 x - 1)}{2 x (3 x + 2) - 1 (3 x + 2)} = \lim_{x 鈫� 1 / 2} \frac{2 (2 x - 1)}{(3 x + 2) (2 x - 1)} = \lim_{x 鈫� 1 / 2} \frac{2}{(3 x + 2)}$

03

Step 3. Let us use the quotient rule in the obtained equation.

$\lim_{x 鈫� 1 / 2} \frac{2}{(3 x + 2)} = \frac{\lim_{x 鈫� 1 / 2} (2)}{\lim_{x 鈫� 1 / 2} (3 x + 2)} = \frac{2}{3 (\frac{1}{2}) + 2} = \frac{2}{\frac{3}{2} + 2} = \frac{2}{\frac{7}{2}} = \frac{2 (2)}{7} = \frac{4}{7}$

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